Consider the parallel RLC circuit shown in Fig. Fig. 1 for $t \ge 0$, driven by a current source $I(t)$. We include the source to keep the formulation general, however, the same derivation applies to the source-free case by simply taking $I(t)=0$ for $t\ge 0$.

To find the governing differential equation for this circuit, we can apply Kirchhoff's current law $$\begin{equation}I(t) = i_R(t) + i_L(t) + i_C(t)\end{equation}$$ where $i_R(t)$, $i_L(t)$, and $i_C(t)$ are the currents through the resistor, inductor, and capacitor, respectively. Using the constitutive relations for these elements, we can write the above equation as $$\begin{equation}I(t) = \frac{v(t)}{R} + C \frac{dv(t)}{dt} + \frac{1}{L} \int_0^t v(t) dt\end{equation}$$ where $v(t)$ is the voltage across the parallel RLC circuit. Differentiating both sides of the above equation with respect to time, we obtain $$\begin{equation}\frac{dI(t)}{dt} = \frac{1}{R} \frac{dv(t)}{dt} + C \frac{d^2v(t)}{dt^2} + \frac{1}{L} v(t)\end{equation}$$ Dividing both sides by $C$ and rearranging the terms, we obtain $$\begin{equation}\frac{d^2v(t)}{dt^2} + \frac{1}{RC} \frac{dv(t)}{dt} + \frac{1}{LC} v(t) = \frac{1}{C} \frac{dI(t)}{dt}\end{equation}$$ This is a second-order linear differential equation with constant coefficients which describes the behavior of the circuit. Comparing it with the general form of the second-order differential equation, we identify $$\begin{equation} \alpha = \frac{1}{2RC}, \quad \omega_0 = \frac{1}{\sqrt{LC}}, \quad f(t) = \frac{1}{C} \frac{dI(t)}{dt} \end{equation}$$ To solve the above differential equation, we first need to find the natural response $v_n(t)$ by solving the corresponding homogeneous equation. Then, if the right hand side is not equal to zero, we need to find the particular solution $v_p(t)$ depending on the form of the forcing fucntion $f(t)$. Finally, the total response of the circuit is given by $v(t) = v_n(t) + v_p(t)$, where the constants in the natural response are determined by the initial conditions of the circuit.

As discussed in the previous section, the solution of the above differential equation depends on the numerical values of $\alpha$ and $\omega_0$. To simplify the presentation, let us now consider the case when $I(t)=I_0$ is constant (where $I_0$ can also be $0$).

• If $\alpha > \omega_0$, the roots $s_1$ and $s_2$ are real and distinct. In this case, the natural response is overdamped and is given by $$\begin{equation}v(t) = A_1 e^{s_1 t} + A_2 e^{s_2 t}\end{equation}$$ where $s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2}$.
• If $\alpha = \omega_0$, the roots $s_1$ and $s_2$ are real and equal. In this case, the natural response is critically damped and is given by $$\begin{equation}v(t) = (A_1 + A_2 t) e^{-\alpha t}\end{equation}$$
• If $\alpha \lt \omega_0$, the roots $s_1$ and $s_2$ are complex conjugates. In this case, the natural response is underdamped and is given by $$\begin{equation}v(t) = e^{-\alpha t} \left[ A_1 \cos(\omega_d t) + A_2 \sin(\omega_d t) \right]\end{equation}$$ where $\omega_d=\sqrt{\omega_0^2-\alpha^2}$.

To find the integration constants we need to use two initial conditions. These initial conditions are coming from the initial voltage across the capacitor and current going through the inductor (at $t=0^+$), which are equal to their values at $t=0^-$ (remember that the voltage across the capacitor and the current through the inductor should be continuous functions). $$\begin{equation} v(0^+) = v(0^-) \end{equation}$$ $$\begin{equation} \frac{dv(t)}{dt}\Big|_{t=0^+} = \frac{1}{C} \left[ I_0 - \frac{v(0^-)}{R} - i_L(0^-) \right] \end{equation}$$ where $v(0^-)$ is the voltage across the capacitor and $i_L(0^-)$ is the current through the inductor just before the perturbation. We can write this conditions because the voltage across a capacitor and the current through an inductor cannot change instantaneously.

The currents going through the resistor and capacitor can be computed easily if you have already computed $v(t)$: $$\begin{equation}i_R(t) = \frac{v(t)}{R}\end{equation}$$ $$\begin{equation}i_C(t) = C \frac{dv(t)}{dt}\end{equation}$$ To compute the current through the inductor, you can use either of the following equations: $$\begin{equation}i_L(t) = I(t) - i_R(t) - i_C(t)\end{equation}$$ $$\begin{equation}i_L(t) = i_L(0^-) + \frac{1}{L} \int_0^t v(t) dt\end{equation}$$

It is apparent from the previous discussion that any voltage and current in the circuit can be expressed in terms of exponentials and sin and cos functions. Therefore, when solving second-order transient problems, it might be more convenient to directly look for the general solution described above. For instance, assuming again the case of current source is constant and equal to $I_0$ for $t\ge 0$, the current going through the inductor can be derived as follows.

• If $\alpha > \omega_0$, the roots $s_1$ and $s_2$ are real and distinct. In this case, the natural response is overdamped and is given by $$\begin{equation}i_L(t) = I_0 + A_1 e^{s_1 t} + A_2 e^{s_2 t}\end{equation}$$ where $s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2}$.
• If $\alpha = \omega_0$, the roots $s_1$ and $s_2$ are real and equal. In this case, the natural response is critically damped and is given by $$\begin{equation}i_L(t) = I_0 + (A_1 + A_2 t) e^{-\alpha t}\end{equation}$$
• If $\alpha \lt \omega_0$, the roots $s_1$ and $s_2$ are complex conjugates. In this case, the natural response is underdamped and is given by $$\begin{equation}i_L(t) = I_0 + e^{-\alpha t} \left[ A_1 \cos(\omega_d t) + A_2 \sin(\omega_d t) \right]\end{equation}$$ where $\omega_d=\sqrt{\omega_0^2-\alpha^2}$.

The integration constants should be found from the initial conditions, we need to analyze the circuit at $t=0^-$ and apply the appropriate initial conditions for capacitors and inductors. At $t=0^+$, we can use the following initial conditions which can be derived by imposing the continuity of the current and voltage in the capacitor and inductor, respectively: $$\begin{equation} i_L(0^+) = i_L(0^-) \end{equation}$$ $$\begin{equation} L\frac{di_L(t)}{dt}\Big|_{t=0^+} = v_C(0^-) \end{equation}$$ where $v_C(0^-)$ is the voltage across the capacitor and $i_L(0^-)$ is the current through the inductor at $t=0^-$.

It is clear that one can apply the same method to compute the other voltages and currents in the circuit, however, this method is easier only when computing the voltage across the capacitor and current through the inductor because, in this cases, one can write one of the two initial conditions relatively easy (it is simply the continuity of the current through the inductor and voltage across the capacitor).